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Let $A\in M_{n}(C)$ be a matrix such that $A^*=-A$ and $A^4=I$.

I need to prove that the eigenvalues of A are between $-i$ to $i$ and that $A^2+I=0$

I didn't get to any smart conclusion.

Thanks

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    Hint for the first part: if z is an eigenvalue of A, show that z^4=1. – Did Aug 11 '11 at 16:17
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    Since $A^*=-A$, i.e. $A$ is skew-hermitian, you can diagonalize. Try that and see what you get. – John M Aug 11 '11 at 16:25
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    What does "between $−i$ to $i$" mean? – lhf Aug 11 '11 at 19:34
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    "Between" may be translating, from a language different from English, words that mean "in" or "among" [the two elements of a 2-element set]. – zyx Aug 12 '11 at 05:53
  • Can't you say in english "all the numbers between this value to another". I believe you can.. –  Aug 12 '11 at 15:16
  • @John M :Can you extend your hint please? I didn't understand much. –  Aug 12 '11 at 15:24
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    One can (provided one uses between this and that instead of between this to that), but only for numbers in a totally ordered set such as the real line, and hardly for the complex plane. Or you wish to speak of the set ${t\mathrm{i}\mid t\ \text{real}, -1\le t\le 1}$, then I believe you can call it the segment between $\mathrm{-i}$ and $\mathrm{i}$. – Did Aug 12 '11 at 18:53

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Do you recall that hermitian matrices ($A^*=A$) must have real eigenvalues? Similiarly, skew-hermitian matrices, i.e. $A^*=-A$, must have pure imaginary eigenvalues. (see Why are all nonzero eigenvalues of the skew-symmetric matrices pure imaginary?)

Also, since $A$ is skew-hermitian, then $A$ is normal too, i.e. $A^*A=AA^*$, so we can apply the spectral theorem: there exists a unitary matrix $U$ such that $A=UDU^{-1}$, where $D$ is a diagonal matrix, whose diagonal entries are the eigenvalues of $A$.

Thus we know that $A^4=(UDU^{-1})^4=UD^4U^{-1}=I$, so $D^4=I$, so all the eigenvalues are 4th roots of unity, i.e. $1,-1,i,\text{ or} -i$. But we already know the eigenvalues are pure imaginary, so all the eigenvalues are $i$ or $-i$. So $D^2=-I$.

Finally, we have $A^2=(UDU^{-1})^2=UD^2U^{-1}=U(-I)U^{-1}=-I$, i.e. $A^2+I=0$.

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  • This theorem can be applied only when A is normal? Is this the only mandatory condition? –  Aug 12 '11 at 18:11
  • Yes, in $M_n(\mathbb{C})$, $A$ diagonalizes in a orthonormal basis $\Leftrightarrow$ $A$ is normal. – Najib Idrissi Aug 12 '11 at 18:19